Here is the correct solution:

First we must prove another theorem:

The reciprocals of two given unequal positive numbers are
positive numbers unequal in the opposite order from the 
two given positive numbers.

[Maybe you've already proved this, but in case you haven't,
here is the proof of that]:

Proof:

Suppose we are given:

0 < x < y  and we are to prove that 0 <  < 

Starting with

0 < x < y

divide all three sides by y:

 <  < 

Simplify:

0 <  < 1

Divide all three sides by x:

 < ÷ < 
 
0 < ÷ <  

0 < · < 

0 < · < 

0 <  < 

----------------------------------

Now that we have proved that, we can turn to your
problem:

4 < a < 7 < b < 9

We are given 

      4 < a  

Divide both sides by 9

(1)    <  

We are given 

      b < 9  

By the theorem above:

      < 

Multiply both sides by "a":

(2)    < 

 
We are given 

      7 < b  

By the theorem above:

      < 

Multiply both sides by "a":

(3)    < 

We are given 

      a < 7  

Divide both sides by 7

(4)    <  

       < 1

Putting (1), (2), (3) and (4) together:

       <  <  < 1

So omit  and we have

       <  < 1

as the only one of the inequalities given that 
we can be sure of.

Edwin